Question

A golfer gives a ball a maximum initial speed of 51.5 m/s. how far does it go

Answer 1

Answer:

Golf ball will go a maximum of 270.36 meter.

Explanation:

  Projectile motion has two types of motion Horizontal and Vertical motion.

 Vertical motion:

          We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

          Considering upward vertical motion of projectile.

          In this case, Initial velocity = vertical component of velocity = u sin θ, acceleration = acceleration due to gravity = -g $m/s^2$ and final velocity = 0 m/s.

         0 = u sin θ - gt

          t = u sin θ/g

     Total time for vertical motion is two times time taken for upward vertical motion of projectile.

     So total travel time of projectile = 2u sin θ/g

Horizontal motion:

   We have equation of motion , $s= ut+\frac{1}{2} at^2$, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

   In this case Initial velocity = horizontal component of velocity = u cos θ, acceleration = 0 $m/s^2$ and time taken = 2u sin θ /g

  So range of projectile,  $R=ucos\theta*\frac{2u sin\theta}{g} = \frac{u^2sin2\theta}{g}$

  Now in the given problem

     A golfer gives a ball a maximum initial speed of 51.5 m/s. how far does it go

     u = 51.5 m/s, for maximum range θ = 45⁰

   So maximum distance reached = $\frac{51.5^2sin(2*45)}{9.81}=270.36 meter$

So it will go a maximum of 270.36 meter.